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How would you prove that every odd integer is a difference of two squares?

I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.

So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it. Could someone help me out please, thanks.

5 Answers5

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Hint:

Compute $(n+1)^2-n^2$.

You should get $2n+1$

Mason
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HINT

$$(k+1)^2-k^2=2k+1$$ $$k^2-(k+1)^2=-(2k+1)$$

Rhys Hughes
  • 12,842
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Hint: What is the difference of squares of two consecutive numbers?

Vasily Mitch
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Hint: As you mentioned, you have $k = 2l+1$ for some $l$.

From here, note the difference of squares:

$$a^2-b^2 = (a+b)(a-b)$$

Let $a$ and $b$ be two consecutive integers and try to simplify.

KM101
  • 7,176
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$$3=4-1$$

$$2n+1=(n+1)^2-n^2$$